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I'm working in additive theory, mainly in the context of non-abelian [semi]groups. I got a PhD in pure mathematics in Nov 2013 from the Université Lyon 1, under the advisorship of François Hennecart and Alain Plagne. As from Jan 2014, I am a post-doc at the Centre de mathématiques Laurent Schwartz.


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comment If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
@Chris: Thanks for your clarifications, which change many things and prove that also my 2nd reading of your answer was incorrect. To be honest, it wasn't so clear (to me?) which open sets you were actually adding to $\tau_X$: there's not a unique way to do it, and the "cheapest" way to do it is to add just $M$. Further, your answer still says: "Let $X$ be any topological space. [...]" So it's not really a question of being complicated or not. Btw, there's a mistake in my last comment: it should really read as "$d(0,x):=d(1,x):=0$ for all $x\in M$ and $d(x,0):=d(x,1):=\infty$ for $x\in X$".
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comment If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
Here's a "natural" follow up: mathoverflow.net/questions/163559.
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revised First-countable topological monoids without local absorbing elements whose topology is induced by a semimetric
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asked First-countable topological monoids without local absorbing elements whose topology is induced by a semimetric
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comment If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
If my 2nd reading of this answer (based on Eric's comments) is right, then Chris' construction doesn't however work, at least in general (and it's not clear to me if it works at all). Eric is in fact right: a semimetric $d_X$ on $X$ always extends to a semimetric $d$ on $M$ by $d(0,x):=d(x,0):=d(1,y):=d(y,1):=\infty$ for all $x,y\in M$ with $x\ne 0$ and $y\ne 1$; but the only neighborhood of both $0$ and $1$ in the canonical topology of $d$ is $M$, so $d$ induces $\tau_X\cup\{M\}$ if $d_X$ induces $\tau_X$ (the topology initially given on $X$). The rest follows from the previous comment.
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comment If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
@Eric Wofsey. I see! So I was misreading Chris' answer. He's just starting with a topology $\tau_X$ on $X$ and taking $\tau=\tau_X\cup\{M\}$ as a topology on $M$, right? That's fine! Still, how to prove that $\tau$ is not induced by a left/right $\mathbb M$-invariant semimetric? This boils down to $\tau$ not being semimetrizable at all (which is close to your remarks), since for a semimetric $d$ on $M$ we would have (for Chris' construction) that, for all $x,y,z\in M$, $d(xz,yz)=d(zx,zy)=d(0,0)\le d(x,y)$ if $z\ne 1$ or $d(xz,yz)=d(zx,zy)=d(x,y)$ if $z=1$, i.e. $d$ would be subinvariant.
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comment If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
(...) for which $\mathbb M$ is the unitization of a null sgrp, how do you think to prove that there doesn't exist any left/right $\mathbb M$-subinvariant semimetric $d$ on $M$ such that $\tau$ the canonical topology induced by $d$? That's the thing.
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comment If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
(...) $(\mathbb M,\tau)$ is not a topological monoid, unless $\{S\}\in\tau$, which is the case iff $S=\{0\}$. In fact, $M\setminus \{0\}$ is an open neighborhood of $1$ in $\tau$ (by construction), and given any open neighborhood $U$ of $(S,S)$ in the product topology induced on $M\times M$ by $\tau$ we should have that $xy\ne 0$ for all $x,y\in U$, but this is false unless we can take $U=\{(S,S)\}$, which in turn is possible iff $M\setminus \{0\}=\{S\}$, i.e. $S=\{0\}$. Do I miss something? And in any case, even assuming that $(\mathbb M,\tau)$ is a 1st-countable topological monoid (...)
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comment If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
[...] the discrete topology on $S$, as far as $S$ is countable, or more generally the topology $\{\emptyset, S\setminus \{0\}, S\}$); it is clear that $(\mathbb S,\tau_S)$ is a topological sgrp. Next, take $\mathbb M$ to be the unitization of $\mathbb S$, so that $\mathbb =S\cup\{S\}$ (I do everything in TG) and $\cdot$ is, by abuse of notation, the unique extension of the composition law of $\mathbb S$ to a binary operation on $M$ for which $xS=Sx=x$ for all $x$. Lastly, let $\tau=\tau_S\cup\{M,M\setminus\{0\}\}$. Then $\tau$ is a 1st-countable topology on $M$; however, (...)
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comment If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
Thank you, Eric, I will keep it in mind. It was yesterday that I discussed the problem with Jacek. We didn't get an answer, but he made a cute observation on the multiplicative structure of the real field, and this morning I ended up with the above.
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comment If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
I had thought of something in the same lines. Yet, I don't see how this answers my question. Essentially, you're taking as $\mathbb M$ the (forced) unitization of a null sgrp. Then, you're claiming that any topology on $M$ is compatible with the structure of $\mathbb M$. But I'm afraid this is false, even assuming that $\tau$ is 1st-countable. To see why, let $\mathbb S$ be a null sgrp with $\mathbb S=(S, \cdot)$, and let $\tau_S$ be any 1st-countable topology on $S$ such that $S \setminus \{0\} \in \tau_S$, where $0$ is the absorbing element of $\mathbb S$ (e.g., $\tau_S$ can be [...]
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revised If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
Wrong adverbs
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revised If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
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revised If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
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answered If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
Apr
13
comment If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
Eric: Thanks for mentioning the example with the Alexandrov topology. As a minor addendum to what you said in your 1st comment, the canonical topology of a semimetric $d$ is T1 if and only if $d(x,y)\ne 0$ for all distinct $x,y$. @Wlodzimierz Holsztynski: I will give a look at Archangielski's work (thanks for the hint). And I agree with the splitting that you suggest.
Apr
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comment If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
I don't expect this to be true in general: let me clarify this point in the OP. (For the record, I've in mind some topological monoids considered by J. Snellman in the context of factorization theory.)
Apr
13
revised If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
Some clarifications after comments of Mariano Suárez-Alvarez
Apr
12
comment If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
True. So let us restrict to "sufficiently small" topological monoids. What happens?
Apr
12
revised If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
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